# If the sections of a function are $\mathscr{L}^2$, would the function be jointly $\mathscr{L}^2$?

Let $(\varOmega,\mathcal{A},\mu)$ and $(\varXi,\mathcal{B},\nu)$ be $\sigma$-finite measure spaces, and $f\colon\varOmega\times\varXi\to\mathbb{R}$ be $\big(\mathcal{A}\otimes\mathcal{B}\big)$-$\mathcal{B}(\mathbb{R})$-measurable. Suppose also that the sections of $f$ are $\mathscr{L}^2$; i.e., $$f(\cdot,\xi)\in\mathscr{L}^2(\varOmega,\mu),~\text{for \nu-a.e.}~\xi,~\text{and}~f(\omega,\cdot)\in\mathscr{L}^2(\varXi,\nu),~\text{for \mu-a.e.}~\omega.$$ Then, is it true that $f\in\mathscr{L}^2(\varOmega\times\varXi,\mu\otimes\nu)$? If not, what restrictions do we need?

The answer is no. You cannot conclude that $f\in\mathscr{L}^2(\varOmega\times\varXi,\mu\otimes\nu)$, see this essentially the same question.
One sufficient condition for the claim to hold though is that $f$ be a tensor product: $f(\omega, \xi) = f_1(\omega)f_2(\xi)$. Since the $\mathscr{L}^p$ play nice with tensor products: $$\|f_1 \otimes f_2\|_{\mathscr{L}^p(\varOmega\times\varXi,\mu\otimes\nu)} = \|f_1\|_{\mathscr{L}^p(\varOmega,\mu)} \|f_2\|_{\mathscr{L}^p(\varXi,\nu)}.$$
• A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(\cdot,\xi)\in\mathcal{L}^2(\Omega,\mu)$ for $\nu$-almost every $\xi$, let $g(\xi)$ be $\|f(\cdot,\xi)\|_{\mathcal{L}^2(\Omega,\mu)}$ (which has just been seen to be $\nu$-almost everywhere finite) and demand that $g\in\mathcal{L}^2(\Xi,\nu)$. That $f(\omega,\cdot)\in\mathcal{L}^2(\Xi,\nu)$ for $\mu$-almost every $\omega$ then follows by Tonelli's Theorem, in addition to $f\in\mathcal{L}^2(\Omega\times\Xi,\mu\otimes\nu)$ as desired. Sep 10 '18 at 20:49